Magnetic field, $B=0.75 \mathrm{~T}$
Accelerating voltage, $\mathrm{V}=15 \mathrm{kV}=15 \times 10^{3} \mathrm{~V}$
Electrostatic field, $E=9.0 \times 10^{-5} \mathrm{~V} \mathrm{~m}^{-1}$
Kinetic energy of the electron, $E=(1 / 2) \mathrm{mv}^{2}$
$\mathrm{eV}=(1 / 2) \mathrm{mv}^{2}$
Therefore, $(e / m)=\left(v^{2} / 2 V\right)$
Here,
Mass of the electron $=m$
Charge of the electron $=\mathrm{e}$
Velocity of the electron $=v$
The electric and magnetic fields have no effect on the particles. As a result, the force exerted on the particle by the electric field is balanced by the force exerted by the magnetic field.
$e E=e v B$
$\Rightarrow \mathrm{v}=\mathrm{E} / \mathrm{B}$
Therefore, $(1 / 2) \mathrm{m}(\mathrm{E} / \mathrm{B})^{2}=\mathrm{eV}$
$\mathrm{e} / \mathrm{m}=\mathrm{E}^{2} / 2 \mathrm{VB}^{2}$
$=\frac{\left(9.0 \times 10^{5}\right)^{2}}{2 \times 15000 \times(0.75)^{2}}=4.8 \times 10^{7} C / \mathrm{kg}$
The beam contains Deuterium ions or deuterons
Because just the charge-to-mass ratio is established, the response is not unique. Other probable responses are He++, Lit+, and so on.